RStudio Exercise 4: Clustering and classification
Housekeeping
First, install required packages:
install.packages('plotly', repos = "http://cloud.r-project.org")
install.packages('dplyr', repos = "http://cloud.r-project.org")
install.packages('ggplot2', repos = "http://cloud.r-project.org")
install.packages('GGally', repos = "http://cloud.r-project.org")
install.packages("corrplot", repos = "http://cloud.r-project.org")Access the required libraries:
library(dplyr)
library(MASS)
library(ggplot2)
library(GGally)Summary of the data
For this exercise, we are using the Boston dataset, which contains information on housing in Boston Massachusetts. For more information, see this website
We start out by loading the data from the MASS package.
# load the data
data("Boston")Explore the dataset Boston:
str(Boston)## 'data.frame': 506 obs. of 14 variables:
## $ crim : num 0.00632 0.02731 0.02729 0.03237 0.06905 ...
## $ zn : num 18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
## $ indus : num 2.31 7.07 7.07 2.18 2.18 2.18 7.87 7.87 7.87 7.87 ...
## $ chas : int 0 0 0 0 0 0 0 0 0 0 ...
## $ nox : num 0.538 0.469 0.469 0.458 0.458 0.458 0.524 0.524 0.524 0.524 ...
## $ rm : num 6.58 6.42 7.18 7 7.15 ...
## $ age : num 65.2 78.9 61.1 45.8 54.2 58.7 66.6 96.1 100 85.9 ...
## $ dis : num 4.09 4.97 4.97 6.06 6.06 ...
## $ rad : int 1 2 2 3 3 3 5 5 5 5 ...
## $ tax : num 296 242 242 222 222 222 311 311 311 311 ...
## $ ptratio: num 15.3 17.8 17.8 18.7 18.7 18.7 15.2 15.2 15.2 15.2 ...
## $ black : num 397 397 393 395 397 ...
## $ lstat : num 4.98 9.14 4.03 2.94 5.33 ...
## $ medv : num 24 21.6 34.7 33.4 36.2 28.7 22.9 27.1 16.5 18.9 ...Dataset contains 506 observations of 14 variables.
CRIM- per capita crime rate by townZN- proportion of residential land zoned for lots over 25,000 sq.ft.INDUS- proportion of non-retail business acres per town.CHAS- Charles River dummy variable (1 if tract bounds river; 0 otherwise)NOX- nitric oxides concentration (parts per 10 million)RM- average number of rooms per dwellingAGE- proportion of owner-occupied units built prior to 1940DIS- weighted distances to five Boston employment centresRAD- index of accessibility to radial highwaysTAX- full-value property-tax rate per $10,000PTRATIO- pupil-teacher ratio by townBLACK- 1000(Bk - 0.63)^2 where Bk is the proportion of blacks by townLSTAT- % lower status of the populationMEDV- Median value of owner-occupied homes in $1000’s
summary(Boston)## crim zn indus chas
## Min. : 0.00632 Min. : 0.00 Min. : 0.46 Min. :0.00000
## 1st Qu.: 0.08204 1st Qu.: 0.00 1st Qu.: 5.19 1st Qu.:0.00000
## Median : 0.25651 Median : 0.00 Median : 9.69 Median :0.00000
## Mean : 3.61352 Mean : 11.36 Mean :11.14 Mean :0.06917
## 3rd Qu.: 3.67708 3rd Qu.: 12.50 3rd Qu.:18.10 3rd Qu.:0.00000
## Max. :88.97620 Max. :100.00 Max. :27.74 Max. :1.00000
## nox rm age dis
## Min. :0.3850 Min. :3.561 Min. : 2.90 Min. : 1.130
## 1st Qu.:0.4490 1st Qu.:5.886 1st Qu.: 45.02 1st Qu.: 2.100
## Median :0.5380 Median :6.208 Median : 77.50 Median : 3.207
## Mean :0.5547 Mean :6.285 Mean : 68.57 Mean : 3.795
## 3rd Qu.:0.6240 3rd Qu.:6.623 3rd Qu.: 94.08 3rd Qu.: 5.188
## Max. :0.8710 Max. :8.780 Max. :100.00 Max. :12.127
## rad tax ptratio black
## Min. : 1.000 Min. :187.0 Min. :12.60 Min. : 0.32
## 1st Qu.: 4.000 1st Qu.:279.0 1st Qu.:17.40 1st Qu.:375.38
## Median : 5.000 Median :330.0 Median :19.05 Median :391.44
## Mean : 9.549 Mean :408.2 Mean :18.46 Mean :356.67
## 3rd Qu.:24.000 3rd Qu.:666.0 3rd Qu.:20.20 3rd Qu.:396.23
## Max. :24.000 Max. :711.0 Max. :22.00 Max. :396.90
## lstat medv
## Min. : 1.73 Min. : 5.00
## 1st Qu.: 6.95 1st Qu.:17.02
## Median :11.36 Median :21.20
## Mean :12.65 Mean :22.53
## 3rd Qu.:16.95 3rd Qu.:25.00
## Max. :37.97 Max. :50.00We can notice that the variable chas is a dummy variable (either 0 or 1), but other variables are continuous and have varying means and variances.
We can see this more clearly by visualizing the data.
Visualize the data
# use ggplot to explore the data
#ggplot(Boston, aes(x=zn)) + geom_histogram(binwidth=50)
ggpairs(Boston, lower = list(combo = wrap("facethist", bins = 20)))
We can also plot the correlation between variables:
library(corrplot)# calculate the correlation matrix and round it
cor_matrix<-cor(Boston) %>% round(digits = 2)
# visualize the correlation matrix
corrplot(cor_matrix, method="circle", type="upper", cl.pos="b", tl.pos="d", tl.cex = 0.6)
There appears to be strong correlation between crime (crim) and pupil-teacher-ratio (ptratio) and accessibility to highways (rad)
Standardize the data for the linear discrimination analysis (LDA)
boston_scaled <- scale(Boston) # center and standardize variables
# summaries of the scaled variables
summary(boston_scaled)## crim zn indus
## Min. :-0.419367 Min. :-0.48724 Min. :-1.5563
## 1st Qu.:-0.410563 1st Qu.:-0.48724 1st Qu.:-0.8668
## Median :-0.390280 Median :-0.48724 Median :-0.2109
## Mean : 0.000000 Mean : 0.00000 Mean : 0.0000
## 3rd Qu.: 0.007389 3rd Qu.: 0.04872 3rd Qu.: 1.0150
## Max. : 9.924110 Max. : 3.80047 Max. : 2.4202
## chas nox rm age
## Min. :-0.2723 Min. :-1.4644 Min. :-3.8764 Min. :-2.3331
## 1st Qu.:-0.2723 1st Qu.:-0.9121 1st Qu.:-0.5681 1st Qu.:-0.8366
## Median :-0.2723 Median :-0.1441 Median :-0.1084 Median : 0.3171
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.:-0.2723 3rd Qu.: 0.5981 3rd Qu.: 0.4823 3rd Qu.: 0.9059
## Max. : 3.6648 Max. : 2.7296 Max. : 3.5515 Max. : 1.1164
## dis rad tax ptratio
## Min. :-1.2658 Min. :-0.9819 Min. :-1.3127 Min. :-2.7047
## 1st Qu.:-0.8049 1st Qu.:-0.6373 1st Qu.:-0.7668 1st Qu.:-0.4876
## Median :-0.2790 Median :-0.5225 Median :-0.4642 Median : 0.2746
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.6617 3rd Qu.: 1.6596 3rd Qu.: 1.5294 3rd Qu.: 0.8058
## Max. : 3.9566 Max. : 1.6596 Max. : 1.7964 Max. : 1.6372
## black lstat medv
## Min. :-3.9033 Min. :-1.5296 Min. :-1.9063
## 1st Qu.: 0.2049 1st Qu.:-0.7986 1st Qu.:-0.5989
## Median : 0.3808 Median :-0.1811 Median :-0.1449
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.4332 3rd Qu.: 0.6024 3rd Qu.: 0.2683
## Max. : 0.4406 Max. : 3.5453 Max. : 2.9865Looking at the scaled data, we can see that each variable is now centered around zero, i.e. the mean is zero by definition, whereas the mean varied in the original data.
# change the object to data frame
boston_scaled <- as.data.frame(boston_scaled)
## Create a categorical variable of the crime rate in the Boston dataset (from the scaled crime rate).
# Use the quantiles as the break points in the categorical variable.
# summary of the scaled crime rate
summary(boston_scaled$crim)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## -0.419367 -0.410563 -0.390280 0.000000 0.007389 9.924110# create a quantile vector of crim and print it
bins <- quantile(boston_scaled$crim)
bins## 0% 25% 50% 75% 100%
## -0.419366929 -0.410563278 -0.390280295 0.007389247 9.924109610# create a categorical variable 'crime'
crime <- cut(boston_scaled$crim, breaks = bins, include.lowest = TRUE, labels = c("low", "med_low", "med_high", "high"))
# look at the table of the new factor crime
table(crime)## crime
## low med_low med_high high
## 127 126 126 127# remove original crim from the dataset
boston_scaled <- dplyr::select(boston_scaled, -crim)
# add the new categorical value to scaled data
boston_scaled <- data.frame(boston_scaled, crime)
# number of rows in the Boston dataset
n <- nrow(boston_scaled)
# choose randomly 80% of the rows
ind <- sample(n, size = n * 0.8)
# create train set
train <- boston_scaled[ind,]
# create test set
test <- boston_scaled[-ind,]LDA on the train set
First, we fit the linear discriminant analysis on the train set by using the categorical crime rate as the target variable and all the other variables in the dataset as predictor variables.
# fit the LDA model
lda.fit <- lda(crime ~ ., data = train)
# print the lda.fit object
lda.fit## Call:
## lda(crime ~ ., data = train)
##
## Prior probabilities of groups:
## low med_low med_high high
## 0.2623762 0.2376238 0.2524752 0.2475248
##
## Group means:
## zn indus chas nox rm
## low 1.01669560 -0.9213572 -0.12375925 -0.8811159 0.37251231
## med_low -0.08325717 -0.2943403 -0.02626030 -0.5776409 -0.09440747
## med_high -0.38467135 0.1608112 0.26805724 0.3442697 0.07616192
## high -0.48724019 1.0149946 -0.03610305 1.0654765 -0.30072478
## age dis rad tax ptratio
## low -0.9279974 0.9305303 -0.6806695 -0.7195178 -0.44093036
## med_low -0.3488857 0.3748458 -0.5464108 -0.4954872 -0.07905908
## med_high 0.4142751 -0.3341579 -0.4132674 -0.3185538 -0.19682797
## high 0.8055068 -0.8497270 1.6596029 1.5294129 0.80577843
## black lstat medv
## low 0.3767145 -0.74951161 0.45874010
## med_low 0.3157161 -0.16424098 0.03765972
## med_high 0.0753590 0.02274549 0.13639604
## high -0.8743343 0.91646554 -0.70400490
##
## Coefficients of linear discriminants:
## LD1 LD2 LD3
## zn 0.11672740 0.717439041 -0.94743492
## indus 0.02136662 -0.242137748 0.51733807
## chas -0.07871185 -0.026728267 0.01456195
## nox 0.26805631 -0.723924494 -1.54050530
## rm -0.09867890 -0.096302305 -0.11996983
## age 0.26862233 -0.392784783 -0.08947827
## dis -0.10572993 -0.334489876 0.24939160
## rad 3.26047480 1.028063848 0.12328271
## tax 0.14080661 -0.032409449 0.43082543
## ptratio 0.16947958 -0.054511632 -0.36276707
## black -0.15500250 0.009488418 0.09620898
## lstat 0.24937837 -0.340035108 0.31465904
## medv 0.22740349 -0.480955322 -0.17422203
##
## Proportion of trace:
## LD1 LD2 LD3
## 0.9513 0.0380 0.0106We can see that the first linear discriminant (LD1) explains most of the differences between groups, as the proportio of the trace is the very large (0.9563)
We can see this more clearly by plotting the results:
# create the function for lda biplot arrows
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "orange", tex = 0.75, choices = c(1,2)){
heads <- coef(x)
arrows(x0 = 0, y0 = 0,
x1 = myscale * heads[,choices[1]],
y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
text(myscale * heads[,choices], labels = row.names(heads),
cex = tex, col=color, pos=3)
}
# target classes as numeric
classes <- as.numeric(train$crime)
# plot the lda results
plot(lda.fit, dimen = 2, col = classes, pch = classes)
lda.arrows(lda.fit, myscale = 1)
First linear discriminant (LD1) explains the high crime areas, but cannot predict differences between the lower crime neighborhoods. Second linear discriminant (LD2) can explain some differences between with the low crime neighbordhoods, with a lower value of LD2 associated with a higher crime levels.
# save the correct classes from test data
correct_classes <- test$crime
# remove the crime variable from test data
test <- dplyr::select(test, -crime)
summary(test)## zn indus chas
## Min. :-0.4872 Min. :-1.55630 Min. :-0.27233
## 1st Qu.:-0.4872 1st Qu.:-0.79322 1st Qu.:-0.27233
## Median :-0.4872 Median :-0.18028 Median :-0.27233
## Mean :-0.1158 Mean : 0.07861 Mean :-0.07933
## 3rd Qu.:-0.4872 3rd Qu.: 1.01499 3rd Qu.:-0.27233
## Max. : 3.5861 Max. : 2.42017 Max. : 3.66477
## nox rm age dis
## Min. :-1.33499 Min. :-3.8764 Min. :-2.33313 Min. :-1.2446
## 1st Qu.:-0.90350 1st Qu.:-0.7293 1st Qu.:-0.59771 1st Qu.:-0.8844
## Median :-0.09230 Median :-0.2151 Median : 0.43430 Median :-0.5102
## Mean : 0.07048 Mean :-0.0796 Mean : 0.08876 Mean :-0.1526
## 3rd Qu.: 0.64339 3rd Qu.: 0.4830 3rd Qu.: 0.94676 3rd Qu.: 0.5353
## Max. : 2.72965 Max. : 3.4733 Max. : 1.11639 Max. : 2.0034
## rad tax ptratio
## Min. :-0.981871 Min. :-1.3068 Min. :-2.51994
## 1st Qu.:-0.637331 1st Qu.:-0.7980 1st Qu.:-0.48756
## Median :-0.522484 Median :-0.2209 Median : 0.11292
## Mean : 0.007837 Mean : 0.0332 Mean :-0.06052
## 3rd Qu.: 1.659603 3rd Qu.: 1.5294 3rd Qu.: 0.80578
## Max. : 1.659603 Max. : 1.7964 Max. : 1.26768
## black lstat medv
## Min. :-3.8685 Min. :-1.35597 Min. :-1.90634
## 1st Qu.: 0.2809 1st Qu.:-0.81088 1st Qu.:-0.63148
## Median : 0.3995 Median :-0.08095 Median :-0.10686
## Mean : 0.0932 Mean : 0.01224 Mean : 0.04163
## 3rd Qu.: 0.4406 3rd Qu.: 0.49775 3rd Qu.: 0.48300
## Max. : 0.4406 Max. : 3.09715 Max. : 2.98650# predict classes with test data
lda.pred <- predict(lda.fit, newdata = test)
# cross tabulate the results
table(correct = correct_classes, predicted = lda.pred$class)## predicted
## correct low med_low med_high high
## low 10 8 3 0
## med_low 7 12 11 0
## med_high 0 5 18 1
## high 0 0 1 26The results in the above table show that the model predicts high crime areas well, but low, medium-low and medium-high crime areas are more difficult to predict. The model is unable to place low crime areas into the low category, given that it predicts low areas to belong to low and medium-low areas with the same probability.
K-means
#Reload the data from the MASS package
library(MASS)
data('Boston')
#Scale the data
boston_scaled <- scale(Boston)Next, we calculate the distances between variables
# euclidean distance matrix
dist_eu <- dist(boston_scaled)
# look at the summary of the distances
summary(dist_eu)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.1343 3.4625 4.8241 4.9111 6.1863 14.3970# manhattan distance matrix
dist_man <- dist(boston_scaled, method = 'manhattan')
# look at the summary of the distances
summary(dist_man)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.2662 8.4832 12.6090 13.5488 17.7568 48.8618We can see that the Euclidean distances are shorter than the Manhattan distances, because the Euclidean distance calculates the shortest distance between variables, whereas the Manhattan distance measure the distance between variables along the axes at right angles.
Next, we look at the results from k-means clustering with optimal number of clusters
set.seed(123) # Fix random number generator
# determine the number of clusters
k_max <- 10
# calculate the total within sum of squares
twcss <- sapply(1:k_max, function(k){kmeans(boston_scaled, k)$tot.withinss})
# visualize the results
qplot(x = 1:k_max, y = twcss, geom = 'line')
Because, the total within sum-of-squares decreases the fastest between one and two clusters, we perform k-means clustering with two centers.
# k-means clustering
km <-kmeans(boston_scaled, centers = 2)
summary(boston_scaled)## crim zn indus
## Min. :-0.419367 Min. :-0.48724 Min. :-1.5563
## 1st Qu.:-0.410563 1st Qu.:-0.48724 1st Qu.:-0.8668
## Median :-0.390280 Median :-0.48724 Median :-0.2109
## Mean : 0.000000 Mean : 0.00000 Mean : 0.0000
## 3rd Qu.: 0.007389 3rd Qu.: 0.04872 3rd Qu.: 1.0150
## Max. : 9.924110 Max. : 3.80047 Max. : 2.4202
## chas nox rm age
## Min. :-0.2723 Min. :-1.4644 Min. :-3.8764 Min. :-2.3331
## 1st Qu.:-0.2723 1st Qu.:-0.9121 1st Qu.:-0.5681 1st Qu.:-0.8366
## Median :-0.2723 Median :-0.1441 Median :-0.1084 Median : 0.3171
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.:-0.2723 3rd Qu.: 0.5981 3rd Qu.: 0.4823 3rd Qu.: 0.9059
## Max. : 3.6648 Max. : 2.7296 Max. : 3.5515 Max. : 1.1164
## dis rad tax ptratio
## Min. :-1.2658 Min. :-0.9819 Min. :-1.3127 Min. :-2.7047
## 1st Qu.:-0.8049 1st Qu.:-0.6373 1st Qu.:-0.7668 1st Qu.:-0.4876
## Median :-0.2790 Median :-0.5225 Median :-0.4642 Median : 0.2746
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.6617 3rd Qu.: 1.6596 3rd Qu.: 1.5294 3rd Qu.: 0.8058
## Max. : 3.9566 Max. : 1.6596 Max. : 1.7964 Max. : 1.6372
## black lstat medv
## Min. :-3.9033 Min. :-1.5296 Min. :-1.9063
## 1st Qu.: 0.2049 1st Qu.:-0.7986 1st Qu.:-0.5989
## Median : 0.3808 Median :-0.1811 Median :-0.1449
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.4332 3rd Qu.: 0.6024 3rd Qu.: 0.2683
## Max. : 0.4406 Max. : 3.5453 Max. : 2.9865Then we plot the original dataset with the clusters
# plot the original Boston dataset with clusters
pairs(boston_scaled, col = km$cluster)
# k-means clustering
km <-kmeans(boston_scaled, centers = 4)
summary(km)## Length Class Mode
## cluster 506 -none- numeric
## centers 56 -none- numeric
## totss 1 -none- numeric
## withinss 4 -none- numeric
## tot.withinss 1 -none- numeric
## betweenss 1 -none- numeric
## size 4 -none- numeric
## iter 1 -none- numeric
## ifault 1 -none- numeric# plot the Boston dataset with clusters
pairs(boston_scaled[, 1:6], col = km$cluster)
# plot the Boston dataset with clusters
pairs(boston_scaled[, 7:14], col = km$cluster)
Bonus
Next, we perform linear discriminant analysis with the Boston data by using the clusters from the k-means testing as the target variables.
# Read data
data(Boston)
head(Boston)## crim zn indus chas nox rm age dis rad tax ptratio black
## 1 0.00632 18 2.31 0 0.538 6.575 65.2 4.0900 1 296 15.3 396.90
## 2 0.02731 0 7.07 0 0.469 6.421 78.9 4.9671 2 242 17.8 396.90
## 3 0.02729 0 7.07 0 0.469 7.185 61.1 4.9671 2 242 17.8 392.83
## 4 0.03237 0 2.18 0 0.458 6.998 45.8 6.0622 3 222 18.7 394.63
## 5 0.06905 0 2.18 0 0.458 7.147 54.2 6.0622 3 222 18.7 396.90
## 6 0.02985 0 2.18 0 0.458 6.430 58.7 6.0622 3 222 18.7 394.12
## lstat medv
## 1 4.98 24.0
## 2 9.14 21.6
## 3 4.03 34.7
## 4 2.94 33.4
## 5 5.33 36.2
## 6 5.21 28.7# Scale data
boston_scaled <- scale(Boston)
# change the object to data frame
boston_scaled <- as.data.frame(boston_scaled)
# Kmeans with three centers
km1 <-kmeans(boston_scaled, centers = 4)
lda.fit1 <- lda(km1$cluster ~ ., data = boston_scaled)
# target classes as numeric
classes <- as.numeric(km1$cluster)
# plot the lda results
plot(lda.fit1, dimen = 2, col = classes, pch = classes)
lda.arrows(lda.fit1, myscale = 1)
We can see that the linear discriminants are useful at explaining the different clusters along the two dimensions, because the clusters are quite separate from one another.
Super Bonus
Use package plotly to create 3D images
library(plotly)##
## Attaching package: 'plotly'## The following object is masked from 'package:ggplot2':
##
## last_plot## The following object is masked from 'package:MASS':
##
## select## The following object is masked from 'package:stats':
##
## filter## The following object is masked from 'package:graphics':
##
## layoutmodel_predictors <- dplyr::select(train, -crime)
# check the dimensions
dim(model_predictors)## [1] 404 13dim(lda.fit$scaling)## [1] 13 3# matrix multiplication
matrix_product <- as.matrix(model_predictors) %*% lda.fit$scaling
matrix_product <- as.data.frame(matrix_product)plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers')